Tessellate S.T.E.M.S. (2019) - Mathematics - Category B - Set 2 - Objective Problem 5

Algebra Level 4

Find the sum of all the possible values of ( x + y ) (x+y) for which ( x , y ) (x,y) satisfies the following pair of equations.

{ x + 3 x − y x 2 + y 2 = 3 y − x + 3 y x 2 + y 2 = 0 \begin{cases} x + \dfrac{3x-y}{x^2+y^2} = 3 \\ y - \dfrac{x+3y}{x^2+y^2} = 0 \end{cases}

This problem is a part of Tessellate S.T.E.M.S. (2019)

5 2 7 3

Tessellate S.T.E.M.S. Mathematics by Tessellate S.T.E.M.S. Ma… , 26

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1 solution

Patrick Corn
Jan 30, 2019

Let z = x + y i . z = x+yi. Then the two equations are the real and complex parts of the equation z + 3 − i z = 3. z + \frac{3-i}{z} = 3. This becomes z 2 − 3 z + ( 3 − i ) = 0 , z^2 - 3z + (3-i) = 0, and the left side factors as ( z − ( 2 + i ) ) ( z − ( 1 − i ) ) . (z-(2+i))(z-(1-i)). So the two solutions are x + y i = 2 + i , 1 − i , x+yi = 2+i, 1-i, or ( x , y ) = ( 2 , 1 ) , ( 1 , − 1 ) . (x,y) = (2,1), (1,-1). The sum of these values is 3 . \fbox{3}.

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